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Angular Velocity of Merry-go-round

  1. Apr 18, 2005 #1
    The problem reads as follows:

    A runner of mass m=36 kg and running at 2.9 m/s runs and jumps on the rim of a playground merry-go-round which has a moment of inertia of 404 kgm^2 and a radius of 2 m. Assuming the merry-go-round is initially at rest, what is its final angular velocity to three decimal places?

    According to the back of the book, the answer is 0.381 rad/s; however, I can never come up with that answer.

    I have tried the following formulas with no luck:
    mvr = Iw
    1/2mv^2 = 1/2Iw^2 (until I realized kinetic energy would not be conserved.)

    I know it has something to do with conservation of angular momentum; however, I cannot seem to figure out the angular momentum of the runner before he jumps onto the merry-go-round. Any help would be greatly appreciated.
     
  2. jcsd
  3. Apr 18, 2005 #2

    Doc Al

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    [tex]\vec{L} = \vec{r} \times \vec{p}[/tex]
     
  4. Apr 18, 2005 #3

    Doc Al

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    Actually, this is the one you need. The initial angular momentum (mvr) equals the final angular momentum. You'll need to find the rotational inertia (I) of the system (merry-go-round + runner).
     
  5. Apr 18, 2005 #4

    OlderDan

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    The angular momentum is the key. You are right about that. I am assuming the runner is approaching the merry-go-round on a tagent line. His angular velocity at the moment of jumping on is his speed times the radius of the merry-go-round, vr. His angular momentum is mvr. That angular momentum will be conserved.

    The moment of inertia of the runner plus the merry-go-round after impact is the sum of the merry-go-round plus the moment of inertia of the runner, which is mr^2.

    Can you take it from there?
     
  6. Apr 18, 2005 #5
    If L = r * p then I have already tried this as it would be the same as L = mvr. I have tried mvr = Iw and solved for w and got the formula w = mvr/I. When I plug the numbers into this formula I do not get the correct answer. Am I missing something?
     
  7. Apr 18, 2005 #6

    Doc Al

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    What are you using for the moment of inertia?
     
  8. Apr 18, 2005 #7
    I finally got the correct answer. I hate it when you are on the right track, but can't find the missing part of the equation. My missing part of the equation was the moment of inertia of the runner. After using the formula mvr = (Imgr + Ir)w (where Imgr is inertia of merry-go-round and Ir is inertia of runner), I solved for w and was able to finally come up with the correct answer.

    Thanks for your help.
     
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